Inverse trigonometric function arise when we want to calculate the angle from side measurements in triangles. They frequently appear in the solutions of differential equations.

There different notation for the inverse function. let $\sin x$ be a function its inverse can be represented as $\sin^{-1}x$. We also write $\arcsin x$, it means the something as $\sin^{-1}x$.

## Derivatives of Inverse Trig Functions Chart

Derivative of inverse trigonometric functions are given below,
$\frac{d}{dx}$$\arcsin x = \frac{1}{\sqrt{1-x^{2}}} \frac{d}{dx}$$\arccos x$ = $\frac{-1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}$$\arctan x = \frac{1}{1+x^{2}} \frac{d}{dx}$$arccot\ x$ = $\frac{-1}{1+x^{2}}$

$\frac{d}{dx}$$arcsec\ x = \frac{1}{x\sqrt{x^{2}-1}} \frac{d}{dx}$$arccsc\ x$ = $\frac{-1}{x\sqrt{x^{2}-1}}$

$\frac{d}{dx}$$\arcsin u = \frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx} \frac{d}{dx}$$\arccos u$ = $\frac{-1}{\sqrt{1-u^{2}}}\frac{du}{dx}$

$\frac{d}{dx}$$\arctan u = \frac{1}{1+u^{2}}\frac{du}{dx} Observe that the derivatives of \arccos x, arccot\ x and arccsc\ x are the negative of the derivatives of \arcsin x, \arctan x and arcsec\ x respectively. ## How to Solve Inverse Trig Functions The following problems will help us to learn more about Inverse Trig Functions ### Solved Examples Question 1: Differentiate y = \arcsin x + x\sqrt{1-x^{2}} Solution: y = \arcsin x + x\sqrt{1-x^{2}} \frac{dy}{dx} = \frac{1}{\sqrt{1-x^{2}}} + x(\frac{1}{2})(-2x)(1-x^{2})^{\frac{-1}{2}} + \sqrt{1-x^{2}} = \frac{1}{\sqrt{1-x^{2}}} - \frac{x^{2}}{\sqrt{1-x^{2}}} + \sqrt{1-x^{2}} = \sqrt{1-x^{2}} + \sqrt{1-x^{2}} = 2\sqrt{1-x^{2}} Question 2: Find the derivative of the function y = \arcsin (\sqrt{x}) Solution: y = \arcsin (\sqrt{x}) we know that \frac{d}{dx}$$\arcsin u$ = $\frac{1}{\sqrt{1-u^{2}}}\frac{du}{dx}$

$\frac{dy}{dx}$ = $\frac{1}{\sqrt{1-(\sqrt{x})^{2}}}\frac{d(\sqrt{x})}{dx}$

= $\frac{1}{\sqrt{1-x}}$.$\frac{1}{2\sqrt{x}}$

= $\frac{1}{2\sqrt{x-x^{2}}}$

Question 3: Find the derivative of the function $y$ = $(1+x^{4})\arctan (x^{2})$?
Solution:

$y$ = $(1+x^{4})\arctan (x^{2})$

$\frac{dy}{dx}$ = $\arctan (x^{2})$$\frac{d}{dx}(1+x^{4}) + (1+x^{4})\frac{d}{dx}$$\arctan(x^{2})$

= $\arctan (x^{2})4x^{3}$ + $(1+x^{4})$$\frac{1}{1+(x^{2})^{2}}$$2x$

= $\arctan(x^{2})4x^{3}$ + $\frac{1+x^{4}}{1+x^{4}}$$2x = 4x^{3} \arctan(x^{2}) + 2X Question 4: Find the derivative of the function y = \frac{d}{dx}$$(arcsec\ e^{3x})$
Solution:

$y$ = $\frac{d}{dx}$$(arcsec\ e^{3x}) we've \frac{d}{dx}$$arcsec\ u$ = $\frac{1}{u\sqrt{u^{2}-1}}\frac{du}{dx}$

= $3 e^3x$${e^{3x}\sqrt{(e^{3x})^{2}-1}}$

= $\frac{3}{\sqrt{e^{6x}-1}}$